# UARS orbital perturbations

1. Sep 29, 2011

### jna

2. Sep 29, 2011

### BobG

First off, if there were no other orbital perturbations than atmospheric drag, then perigee would stay constant while apogee decreased. The orbit has to circularize before it can actually start decaying further into the atmosphere if atmospheric drag were the only perturbation. That's why apogee is decreasing faster than perigee. In this case, apogee isn't incredibly high, so atmospheric drag would reduce perigee a tiny bit.

That means the gravity of some other object must be affecting it, such as the Sun and Moon. The Sun has the more significant effect.

When the Sun is lined up with perigee, the net force of gravity decreases at perigee (Earth gravity minus Sun gravity), meaning the satellite's current speed at perigee will carry it out further from the Earth - i.e. increase apogee. Meanwhile, the net force of gravity at apogee increases (Earth gravity plus Sun gravity), which means the satellite's current speed at apogee won't carry it as far - i.e. decrease perigee.

When the Sun is lined up with apogee, the opposite occurs.

The cycle won't match up with the year because of several things. The most significant is the equatorial bulge (Earth's oblateness). The bulge will cause perigee to move within the orbital plane and the bulge will also cause regression of your ascending/descending nodes (in other words, the orbit plane itself is rotating around the Earth).

Last edited: Sep 29, 2011
3. Sep 29, 2011

### jna

Thanks. But how does the gravitational view explain absence of these oscillations at 570km?

4. Sep 29, 2011

### BobG

That would be hard to say since I don't know much about their operations.

Seeing as how they were able to put it into a disposal orbit, they definitely did have the capability to perform station keeping maneuvers to maintain their orbit. That would be a decent guess, but only a guess. Many low earth satellites have no capability to change their orbit, so the capability implies intent.

The problem is interpreting what they meant by its scientific life ending in 2005. Sensors (and other electronics) fail at different times and I'm pretty sure the design life for most of them wasn't 14 years. They seem to quit caring as much about its altitude around 2000 or so.

Last edited: Sep 29, 2011
5. Sep 29, 2011

### jna

Interesting - haven't thought of that (station control). That surely makes this all very plausible.

6. Sep 29, 2011

### D H

Staff Emeritus
That's not quite true. A better way to look at it, but still simplistic, is that the drag a vehicle undergoes now reduces the orbital radius half an orbit later. Since the vehicle was still in the atmosphere even at apogee, drag at apogee will reduce perigee. Since the atmosphere thins out drastically with altitude, the apogee will decrease a lot faster than will perigee.

The Moon has a more significant effect than the Sun, by about a factor of two. Third body perturbations are tidal. However, that is not the cause of this sinusoidal behavior. It is caused by the atmosphere's diurnal bulge and how the Earth's equatorial bulge moves the argument of perigee around the orbital plane.

Drag decreases the satellite's semi-major axis and it's eccentricity. Perigee will increase if $\dot e < \frac{\dot a}{a}(1-e)$. So, what can make drag have a greater effect on the satellite's eccentricity than its energy? The answer is where along the orbit the vehicle flies through the atmosphere's diurnal bulge. Drag effects eccentricity more than energy in some orbital configurations, making perigee increase even though semi-major axis is decreasing. Change the argument of perigee by 180 degrees and drag effects energy more than nominal, so now perigee drops (quickly). The Earth's equatorial bulge makes the argument of perigee precess within the orbital plane. The end result is that sinusoidal behavior noted by the OP.

7. Sep 30, 2011

### jna

@D H: your approach (drag-induced precession) is actually close to my initial thinking. Except I had suspected seasonal atmospheric density variations throughout the year. From the chart it seems the oscillations have exactly 2 periods per year, so does the annual atmospheric density curve.

8. Sep 30, 2011

### BobG

Just using SGP4 elsets from 2007 (after the satellite was put in a disposal orbit, but well before decay), you can picture both the short term perturbations and long term perturbations. The elsets are just propagated out, so this is based on models, rather than actual observations, but SGP 4 is a decent model.

The right ascension of ascending node rotates a complete 360 degrees in slightly less than 3 months and the argument of perigee advances a complete 360 degrees in approximately 6 months.

Just graphing eccentricity (which indirectly indicates that something is happening with perigee/apogee), you can see there's a lot of short term variations. But the only thing that shows on the graph in the original post are the long term variations. While D_H is right, I don't think the things he's talking about are showing on a graph that just maps the long term changes. I'd have to think about that awhile, though. I work with a lot of geos (a much quieter environment), but really haven't done much with low earth orbits.

Plus perigee/apogee for a shorter time period, which makes it easier to see the detail.

#### Attached Files:

File size:
38.8 KB
Views:
153
• ###### peri_apo.JPG
File size:
48.2 KB
Views:
133
9. Sep 30, 2011

### D H

Staff Emeritus
And that pretty much explains it. I did not want to go into the Lagrange Planetary Equations in my original response; too much information and without the background it would just be a mathout. (And I still don't want to bother with the details of the math; it would beg too many questions).

That said, there are two terms in de/dt, the time derivative of eccentricity. One term involves da/dt. This coupling between da/dt and de/dt means that eccentricity decays toward zero as the semi major axis decreases. The other term involves the partial of the perturbing force with respect to argument of perigee. It is this second term, coupled with the six month apsidal precession, that drives these oscillations.

10. Sep 30, 2011

### BobG

Just for the heck of it, here's how the eccentricity of geosynchronous satellites would vary if left alone.

Oblateness of the earth has an effect on right ascension of ascending node and argument of perigee, but you're measuring the time for a complete cycle in 1000's of days. You also get precession of the orbit plane because of the Sun & Moon (plus a change in inclination), so the end result is that it takes around 3 years or so for RAAN to rotate 360 degrees. The overall environment is a lot quieter.

As mentioned by D_H, the Moon has a more significant effect and causes steep slopes. The Moon just doesn't have time to do much before it cancels itself out the other direction.

There's no reason that effect would disappear for lower orbits, but the variation in eccentricity for UARS is about 4 to 5 times too big to be just because of the orientation of the orbit relative to the Sun.

#### Attached Files:

• ###### geo_ecc.JPG
File size:
36.8 KB
Views:
136
Last edited: Sep 30, 2011
11. Sep 30, 2011

### D H

Staff Emeritus
Yes, there is. Those third body effects are essential tidal gravitational forces. They are, to first order, on the order of GMr/R3, where M is the mass of the perturbing body (i.e., the moon or the sun), R is the distance from the center of the earth to the center of the perturbing body, and r is the distance from the center of the earth to the satellite. Geosync is at about 6 earth radii while LEO is about 1, so those effects are about six times greater for GEO than LEO -- at that is absolute magnitude. Relative to the dominant earth gravitational acceleration, those tidal effects are more than 200 times greater at GEO than at LEO.

12. Oct 4, 2011

### jna

@D_H: Is there a "mechanistic" way of imagining how drag can at all increase the perigee? By mechanistic I refer to any reasonably imaginable way of this without resorting to heavy duty analysis. For example, I understand how a drag at perigee decreases apogee and the eccentricity (while keeping perigee unchanged). Similarly, I see easily how drag applied at apogee decreases perigee and increases the eccentricity. I am missing a similarly intuitive picture of how drag can actually increase perigee. Is such a picture possible?

Many thanks for your time and help!

13. Oct 5, 2011

### D H

Staff Emeritus
jna,

Let's start with the equation for the radial distance as a function of semi-major axis $a$, eccentricity $e$, and true anomaly $\theta$ for a Keplerian orbit:

$$r = \frac{a(1-e^2)}{1-e\cos\theta}$$

Perigee occurs when $\theta = \pi$:

$$r_p = a(1-e)$$

Differentiating with respect to time,

$$\dot r_p = \dot a(1-e) - a\dot e$$

So the condition for an increasing perigee distance is

$$\dot e < (1-e)\frac{\dot a}{a}$$

Note that from Jan 2006 to Jan 2010 the satellite's semi-major axis decreased by only 10 or so kilometers per year (this was thanks to the prolonged solar minimum between solar cycles 23 and 24). Since the semi major axis is about 6800 km, this slow decline in the semi major makes $\dot a/a$ rather small. The condition for an increasing perigee distance in this time frame reduces to $\dot e < -0.0147/\text{year}$. A quick glance at the eccentricity plot attached with post #8 shows that the eccentricity declined much faster than this.

Without derivation, the instantaneous change in eccentricity due to drag is given by

$$\frac{de}{dt} = -\,\frac{A C_D}{m} \rho v (1-e^2)\frac{\cos E}{1-e\cos E}$$

where e is the eccentricity, A is the satellite cross section area, CD is the coefficient of drag, m is the satellite's mass, [and E is the eccentric anomaly. Averaging this instantaneous value over an orbit or longer yields a more meaningful value.

What we want is this average rate of change in the eccentricity to be negative (strongly negative). This will happen if the values of the expression when cos(E) is positive dominate over the values when cos(E) is negative. This will typically be the case because density tends to decrease exponentially with altitude. Ignoring the diurnal bulge, this is not a strong enough effect to overcome the $\dot a / a$ term. Thanks to the diurnal bulge, if perigee more or less coincides with the bulge the increased density due to the bulge will make $\dot e$ strongly negative. The bulge can also make $\dot e$ positive if perigee and passage through the bulge are 180 degrees out of sync.

The heavy duty stuff is Lagrange's Planetary Equations. For example, see http://ccar.colorado.edu/asen5050/ASEN5050/Lectures_files/lecture15.pdf and http://ccar.colorado.edu/asen5050/ASEN5050/Lectures_files/lecture16.pdf. These lecture notes give a brief overview of the concept of variation of parameters with respect to orbits. Atmospheric drag is a nonconservative force, so you have to use Gauss' form of the Lagrange Planetary Equations. This paper, http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1976CeMec..14..335J&amp;data_type=PDF_HIGH&amp;whole_paper=YES&amp;type=PRINTER&amp;filetype=.pdf [Broken] gives a simple treatment of atmospheric drag; it does not address the diurnal bulge.

Last edited by a moderator: May 5, 2017
14. Oct 5, 2011

### jna

Thanks - this is very helpful!

15. Oct 6, 2011

### jna

I am trying to imagine the action: suppose there is a drag force acting exactly at the perigee. This results in lowering the altitude of every point of the orbit except the perigee, making both $\dot e$ and $\dot a$ negative. I can't explain to myself how any amount of drag applied at perigee could cause the satellite to attain a higher perigee. Where am I failing?

16. Oct 6, 2011

### D H

Staff Emeritus
Where you are failing is that you aren't looking at the whole picture. The whole picture is that the satellite was in low Earth orbit; it was in the Earth's atmosphere at all times. You can't get a picture of what happens as a whole by looking at a single part of the orbit.